Re: [Ifeffit] Cumulant expansion fittings

Matt, Scott: There is a situation in one dimensional world, not so practically useful though, where the third cumulant is zero. It describes the 1st nearest neighbor interaction between a central atom (x) in the group of three atoms: A-----x-----A. Indeed, here the third cumulant is zero, the higher order - may be not, depending on the shape of the pair potential. Assume the pair potential as: V(x) = 1/2*kx^2 + k3*x^3, (1) Then, the asymmetric term vanishes since: V_eff(x) = V(x) + V(-x) = kx^2. (2) It is not obvious that there exist other examples. As I am sure you know, Hung, Hung/Rehr, Yokoyama, Pirog and others have studied behavior of cumulants in the most common structures, including fcc, hcp and bcc, using Morse potential. In all of them, the third cumulants were nonzero (albeit Hung/Rehr article contained an error in the third cumulant calculation, and these results were further generalized by Vila/Rehr in their recent paper. Indeed, the potential in these cases is always a linear combination of crystallography-defined terms, e.g., for the fcc structure, assuming a displacement x along the nearest neighbor distance: V_eff(x) = V(x)+2V(-x/2)+8V(x/4)+8V(-x/4)+4V(0) = 1/2*(5/2)*kx^2 + (3/4)*k3*x^3. (3) If you take above equation for V(x) and plug it in, you will obtain that V_eff(x) is still anharmonic since the x^3 term is still present in the effective contribution. It does not matter in this case, whether the atoms surrounding the dopant are "fixed" or "not fixed", as long as their is an effective pair potential between the central atom and its neighbors described by equation (1). Same results can be obtained for most other lattices. Note that even if for some displacement direction x the sum of the terms in Eq. (3) will have no x^3, there is no chance that it will be true for ALL directions, since there is nothing special in x pointing along the 1NN distance. Thus, I am pretty much convinced, unless there is some mistake in my reasoning, that no case exists in 3D with a zero third cumulant. Anatoly ________________________________ From: ifeffit-bounces@millenia.cars.aps.anl.gov on behalf of Matt Newville Sent: Wed 1/21/2009 1:27 PM To: XAFS Analysis using Ifeffit Subject: Re: [Ifeffit] Cumulant expansion fittings Umesh, Scott, Anatoly, I'm not sure what Scott means by a "lattice atom" in "the limit in which the lattice atoms are fixed in place". Materials have atoms. Crystals and lattice points. But I think I do agree with his approach to constructing a non-harmonic distribution for which the third cumulant is zero. Cumulants are simple combinations of the moments of a distribution that are particularly useful when one has an exponential function of a variable and wants to model a distribution function of that variable. For XAFS, the distribution we care about is r (interatomic distance), which has moments: / = | (r-r_0)^n g(r) dr / where g(r) is the distribution function about a value r_0 (not necessarily the mean value for g(r), but usually close to it). To remind everyone, cumulants are combinations of the moments: C1 = <r> ~= r_0 C2 = - <r>^2 C3 = - 3<r> + 2<r>^3 C4 = - 3^2 - 4<r> + 12<r>^2 - 6 <r>^4 (NB: The Wikipedia page on cumulants has different coefficients for C4. I believe the ones above are as in Kendell's Advanced Theory of Statistics and also what are used at http://mathworld.wolfram.com/Cumulant ) For a normal (aka Gaussian) distribution, the 1st and 2nd cumulants (mean and variance) are non-zero and all higher cumulants are zero. We often call this the "harmonic" case. It is definitely possible to have distributions that are "symmetric" (having a g(r) function such that g(r_0 - x) = g(r_0 + x) ) but not harmonic -- I think that is what Scott is saying. It should also be possible to construct a distribution that has C3 = 0 and a non-zero C4. The real question was "can one use the fourth cumulant without the third cumulant in fitting (XAFS)"? The answer is: Yes. As Scott and Anatoly suggested, doing that may not make a great physical model for g(r), but perhaps Umesh has a good reason to try this. My advice is to try it and see if you get a fourth cumulant that is clearly non-zero. My experience (and all the experiences I've heard of) is that the fourth cumulant rarely matters. This is probably related to the idea that the cumulant expansion diverges for very disordered systems and you would either need many higher order cumulants to describe such a distribution or are much better off using a finite set of atomic distances with some model for the weighting of the different distances. The first option (using many cumulants) is impractical, and probably computationally dangerous (as you'd quickly explore issues with computer precision of floating point numbers). The second option (often called a "histogram" approach in the Ifeffit world, and modeled somewhat after The GNXAS Approach) has been used successfully a number of times. --Matt

On Jan 21, 2009, at 10:12 AM, Frenkel, Anatoly wrote:

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Hi Scott,

Third cumulant in your example will not be zero because this arrangement is symmetric only on the average. Locally, the interatomic pair potential (and the cumulants are the measures of the effective pair potential) which is the sum of the two potentials - between the interestitial and its neighbors on the opposite sides)- is still asymmetric, since the repulsive bruch of the potential is steeper than the attractive brunch. You can model your situation using two anharmonic pair potentials, e.g., Morse potential (see, for example, Rehr-Hung's paper in Phys Rev B in the 1990's, and I've done such calculations too, just in the case you described) and you will obtain that the effective pair potential is still analytically anharmonic and it has a non-zero third cumulant.

Anatoly

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